show that $\sum_{n=1}^{2015}a_{n}\equiv 3\pmod 4$

Question

Assmue that real sequence $\{a_{n}\}$ such $$a_{1}=1,|a_{n+1}|=2|a_{n}|$$

show that $$\sum_{n=1}^{2015}a_{n}\equiv 3\pmod 4$$

I have solve $$|a_{n}|=|a_{1}|\cdot 2^{n-1}=2^{n-1}\Longrightarrow a_{n}=\pm 2^{n-1}$$

Well and now I'm stuck and don't know how to proceed.

score 1 · Accepted Answer · answered May 17 '15 at 15:50

1

Since $a_0 > 0$, then we just have $a_n = 2^{n-1}$. Then notice that for $n>2$:

$2^n$ = $4* 2^{n-2} \equiv 0$ mod $4$.

So your sum = $2^0 + 2^1 + \sum_{n=2}^{2015} a_n = 1 + 2 + 0$ (mod $4$) $= 3$ (mod $4$)

answered May 17 '15 at 15:50

Brenton

1

oh,it's nice.+1 – May 17 '15 at 15:51
What? Why are you saying $a_n=2^{n-1}$? We could also have negative terms which satisfy this... – shalop May 17 '15 at 16:41
Ah yes, you are right. I misread it a little. In that case, you need a second case for -$2^1$, but $2^0$ and the $ \sum_{n=2}^{2015} a_n$ term should mod out to the same thing – Brenton May 17 '15 at 16:53

1 Answers1